Trigonometric Functions and the Unit Circle: practice solutions, Grade 12 – download the PDF

1 Degrees to radians ★★★

Multiply by \(\dfrac{\pi}{180}\).

1. \(45\cdot\dfrac{\pi}{180}=\dfrac{\pi}{4}\)
2. \(120\cdot\dfrac{\pi}{180}=\dfrac{2\pi}{3}\)
3. \(270\cdot\dfrac{\pi}{180}=\dfrac{3\pi}{2}\)
4. \(15\cdot\dfrac{\pi}{180}=\dfrac{\pi}{12}\)

2 Radians to degrees ★★★

Multiply by \(\dfrac{180}{\pi}\).

1. \(\dfrac{5\cdot180}{6}=150^\circ\)
2. \(\dfrac{7\cdot180}{4}=315^\circ\)
3. \(\dfrac{180}{10}=18^\circ\)
4. \(\dfrac{3\cdot180}{5}=108^\circ\)

3 Length of an arc ★★★

Use \(s=r\theta\) with \(\theta\) in radians.

\(s=8\cdot\dfrac{\pi}{4}=2\pi\approx 6.28\text{ m}\).

Answer: the path is \(2\pi\approx 6.28\text{ m}\) long (about \(20.6\) ft).

4 Reading the unit circle ★★★

Use the unit circle points \(\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\), \(\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\), \((0,1)\) and \((-1,0)\).

1. \(\dfrac{1}{2}\)
2. \(\dfrac{1}{2}\)
3. \(\tan\dfrac{\pi}{4}=\dfrac{\sqrt{2}/2}{\sqrt{2}/2}=1\)
4. \(0\)
5. \(0\)

5 Quadrants and signs ★★★

Compare each angle with \(\dfrac{\pi}{2},\pi,\dfrac{3\pi}{2},2\pi\).

1. \(\dfrac{\pi}{2}<\dfrac{5\pi}{6}<\pi\): quadrant II, \(\sin>0\), \(\cos<0\).
2. \(\pi<\dfrac{4\pi}{3}<\dfrac{3\pi}{2}\): quadrant III, \(\sin<0\), \(\cos<0\).
3. \(\dfrac{3\pi}{2}<\dfrac{11\pi}{6}<2\pi\): quadrant IV, \(\sin<0\), \(\cos>0\).
4. \(\dfrac{7\pi}{12}=105^\circ\): quadrant II, \(\sin>0\), \(\cos<0\).

6 Right triangle ratios ★★★

Pythagorean theorem: \(h^{2}=5^{2}+12^{2}=169\), so \(h=13\text{ in}\).

\(\sin\theta=\dfrac{5}{13}\), \(\cos\theta=\dfrac{12}{13}\), \(\tan\theta=\dfrac{5}{12}\).

7 Coterminal angles ★★★

Add or subtract multiples of \(2\pi\).

1. \(\dfrac{13\pi}{6}-2\pi=\dfrac{\pi}{6}\)
2. \(-\dfrac{\pi}{3}+2\pi=\dfrac{5\pi}{3}\)
3. \(\dfrac{9\pi}{4}-2\pi=\dfrac{\pi}{4}\)
4. \(-\dfrac{7\pi}{4}+2\pi=\dfrac{\pi}{4}\)

8 Reference angles and values ★★★

1. Quadrant II, reference \(\pi-\dfrac{5\pi}{6}=\dfrac{\pi}{6}\); sine is positive: \(\dfrac{1}{2}\).
2. Quadrant III, reference \(\dfrac{\pi}{3}\); cosine is negative: \(-\dfrac{1}{2}\).
3. Quadrant IV, reference \(\dfrac{\pi}{4}\); tangent is negative: \(-1\).
4. Quadrant IV, reference \(\dfrac{\pi}{6}\); cosine is positive: \(\dfrac{\sqrt{3}}{2}\).

9 Reciprocal functions ★★★

1. \(\sec\dfrac{\pi}{3}=\dfrac{1}{1/2}=2\)
2. \(\sin\dfrac{5\pi}{6}=\dfrac{1}{2}\), so \(\csc=2\)
3. \(\cot\dfrac{3\pi}{4}=\dfrac{-\sqrt{2}/2}{\sqrt{2}/2}=-1\)
4. \(\cos\dfrac{7\pi}{6}=-\dfrac{\sqrt{3}}{2}\), so \(\sec=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}\)

10 All six from one value ★★★

\(\cos^{2}\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\). In quadrant II cosine is negative, so \(\cos\theta=-\dfrac{4}{5}\).

\(\tan\theta=\dfrac{3/5}{-4/5}=-\dfrac{3}{4}\), \(\ \csc\theta=\dfrac{5}{3}\), \(\ \sec\theta=-\dfrac{5}{4}\), \(\ \cot\theta=-\dfrac{4}{3}\).

11 A leaning ladder ★★★

The ladder is the hypotenuse.

Height: \(20\sin70^\circ\approx 20(0.9397)\approx 18.79\text{ ft}\).

Distance: \(20\cos70^\circ\approx 20(0.3420)\approx 6.84\text{ ft}\).

Answer: it reaches about \(18.79\) ft up the wall, and its foot is about \(6.84\) ft from the wall.

12 Sector of a pizza ★★★

Convert: \(100^\circ=\dfrac{5\pi}{9}\).

Arc: \(s=15\cdot\dfrac{5\pi}{9}=\dfrac{25\pi}{3}\approx 26.18\text{ in}\).

Area: \(A=\tfrac{1}{2}(15)^{2}\cdot\dfrac{5\pi}{9}=\dfrac{125\pi}{2}=62.5\pi\approx 196.35\text{ in}^{2}\).

13 True or false? ★★★

1. True: sine is odd.
2. True: cosine is even.
3. True: tangent has period \(\pi\) (the point at \(x+\pi\) is opposite, so sine and cosine both change sign and the ratio is unchanged).
4. False: \(\sin(x+\pi)=-\sin x\). For \(x=\dfrac{\pi}{2}\): \(\sin\dfrac{3\pi}{2}=-1\) while \(\sin\dfrac{\pi}{2}=1\).

14 A bicycle wheel ★★★

\(\omega=45\cdot\dfrac{2\pi}{60}=\dfrac{3\pi}{2}\approx 4.71\text{ rad/s}\).

\(v=r\omega=13\cdot\dfrac{3\pi}{2}=19.5\pi\approx 61.26\text{ in/s}\).

In mph: \(61.26\cdot\dfrac{3600}{63{,}360}\approx 3.48\text{ mph}\).

15 Solving sin equals a value ★★★

Sine is negative in quadrants III and IV. The reference angle satisfies \(\sin\alpha=\dfrac{\sqrt{3}}{2}\), so \(\alpha=\dfrac{\pi}{3}\).

Quadrant III: \(\theta=\pi+\dfrac{\pi}{3}=\dfrac{4\pi}{3}\). Quadrant IV: \(\theta=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}\).

Answer: \(\theta=\dfrac{4\pi}{3}\) or \(\dfrac{5\pi}{3}\).

16 Solving a tangent equation ★★★

Tangent is negative in quadrants II and IV. The reference angle has \(\tan\alpha=\sqrt{3}\), so \(\alpha=\dfrac{\pi}{3}\).

Quadrant II: \(\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\). Quadrant IV: \(2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}\).

Answer: \(\theta=\dfrac{2\pi}{3}\) or \(\dfrac{5\pi}{3}\).

17 Large and negative angles ★★★

1. \(-\dfrac{11\pi}{6}+2\pi=\dfrac{\pi}{6}\), so the value is \(\dfrac{1}{2}\).
2. \(\dfrac{25\pi}{4}-6\pi=\dfrac{\pi}{4}\), so the value is \(\dfrac{\sqrt{2}}{2}\).
3. The period of tangent is \(\pi\): \(\dfrac{17\pi}{3}-5\pi=\dfrac{2\pi}{3}\), and \(\tan\dfrac{2\pi}{3}=-\sqrt{3}\).
4. \(-\dfrac{13\pi}{3}+4\pi=-\dfrac{\pi}{3}\); \(\cos\left(-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\), so \(\sec=2\).

18 Finding sine and tangent from cosine ★★★

Cosine negative and tangent positive means sine is negative too: quadrant III.

\(\sin^{2}\theta=1-\dfrac{25}{169}=\dfrac{144}{169}\), so \(\sin\theta=-\dfrac{12}{13}\).

\(\tan\theta=\dfrac{-12/13}{-5/13}=\dfrac{12}{5}\) and \(\csc\theta=-\dfrac{13}{12}\).

19 The Ferris wheel ★★★

\(\omega=\dfrac{2\pi}{24}=\dfrac{\pi}{12}\approx 0.26\text{ rad/min}\).

The radius is \(75\text{ ft}\): \(v=75\cdot\dfrac{\pi}{12}=\dfrac{25\pi}{4}\approx 19.63\text{ ft/min}\).

\(19.63\cdot\dfrac{60}{5280}\approx 0.22\text{ mph}\). The ride is gentle.

20 Height of a tree ★★★

The side from her eye to the horizontal at the treetop level: \(50\tan36^\circ\approx 50(0.7265)\approx 36.3\text{ ft}\).

Add her eye height: \(36.3+5.5\approx 41.8\text{ ft}\).

Answer: the tree is about \(41.8\) ft tall (about \(12.7\) m).

21 Using an identity ★★★

\(\sec^{2}\theta=1+\tan^{2}\theta=5\). Cosine (so secant) is negative in quadrant III: \(\sec\theta=-\sqrt{5}\).

\(\cos\theta=-\dfrac{1}{\sqrt{5}}=-\dfrac{\sqrt{5}}{5}\) and \(\sin\theta=\tan\theta\cos\theta=-\dfrac{2\sqrt{5}}{5}\).

Check: \(\left(\dfrac{2\sqrt{5}}{5}\right)^{2}=\dfrac{4}{5}\) and \(\left(\dfrac{\sqrt{5}}{5}\right)^{2}=\dfrac{1}{5}\), and \(\dfrac{4}{5}+\dfrac{1}{5}=1\).

22 A car tire ★★★

Speed: \(60\text{ mph}=\dfrac{60\cdot63{,}360}{3600}=1056\text{ in/s}\). Radius: \(14\text{ in}\).

\(\omega=\dfrac{v}{r}=\dfrac{1056}{14}\approx 75.43\text{ rad/s}\).

In rpm: \(75.43\cdot\dfrac{60}{2\pi}\approx 720\text{ rpm}\).